C Language function called with few arguments, but not error?

Question

In the following C code, function printr called only with one parameter, But all compiled with no warning on GCC and VS.
I am confused why this is OK? Thanks!

#include <stdlib.h>
#include <stdio.h>

int printr(i, j){
    printf("The first parameter %d\n", i);
    printf("The second parameter %d\n", j);
    return 0;
}

int main(){
    printr(3);
    return 0;
}

Try using a prototype: `int printr(int i, int j) { /* ... */ }`. You are using old-stye function definition — pmg, Apr 17 '14 at 12:32
It's a duplicate, I've seen this question before, but I don't have a link. — sashoalm, Apr 17 '14 at 12:33
@pmg This looks rather like implicit int; old-style would be empty parentheses instead of void or somesuch. — Jens, Apr 17 '14 at 12:42

score 3 · Answer 1 · answered Apr 17 '14 at 12:37

Gcc does warn you:

$ gcc -Wextra y.c
y.c: In function ‘printr’:
y.c:4: warning: type of ‘i’ defaults to ‘int’
y.c:4: warning: type of ‘j’ defaults to ‘int’

And once you've fixed those it will warn

y.c: In function ‘main’:
y.c:11: error: too few arguments to function ‘printr’

score 2 · Accepted Answer · answered Apr 17 '14 at 12:36

2

You defined printr() by using old-fashion function definition syntax, therefore compiler cannot do some syntactic check. You should define it like this:

int printr(int i, int j) {

By the way, with -Wextra, gcc will give warnings about your definition.

answered Apr 17 '14 at 12:36

Lee Duhem

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C Language function called with few arguments, but not error?

2 Answers2